Equation of a Tangent Line to a Curve: Derivative Functions - Mathematics (Grade 12)

Relationship Between Derivatives and Tangent Lines

You probably already know that derivatives can tell us a lot about the properties of a function. One of the coolest uses of derivatives is to find the slope or gradient of a tangent line to a curve.

Imagine "zooming in" on a point on a curve over and over. Eventually, the curved line will start to look like a straight line, right? Well, that imaginary straight line is what we call the tangent line. The gradient of the tangent line at a point $x=a$ on the curve $y=f(x)$ is exactly equal to the derivative of the function at that point, which is $m = f'(a)$ .

Determining the Equation of a Tangent Line

To create the equation of a straight line, we need two main things: a point the line passes through and the gradient of the line itself. In this context:

Point of Tangency: A point $P(a, b)$ where the line touches the curve.
Gradient (m): The slope of the line at that point, which we get from the derivative, $m = f'(a)$ .

Once we have both, we can plug them directly into the basic formula for a line equation that you're already familiar with:

y - b = m(x - a)

In short, to find the equation of a tangent line, first find its gradient by differentiating the function, then plug the point of tangency and the gradient into the line equation formula.

Breaking Down a Case

Let's see how it works with an example. Determine the equation of the tangent line to the parabola $y = 2 - 4x^2$ at the point $(1, -2)$ .

Solution:

Step $1$ : Find the gradient of the tangent line

First, we differentiate the function $f(x) = 2 - 4x^2$ to get its gradient expression.

f'(x) = -8x

Since we want to find the gradient at the point where the $x$ -coordinate is $x=1$ , we substitute this value into the derivative.

m = f'(1) = -8(1) = -8

So, the gradient of the tangent line is $-8$ .

Step $2$ : Construct the equation

Now we have everything we need:

Point of tangency $(a, b) = (1, -2)$
Gradient $m = -8$

Plug them into the line equation formula:

y - b = m(x - a)

So, the equation of the tangent line is $y = -8x + 6$ .

Visualization of the Tangent Line on the Parabola

This graph shows the parabola curve

y = 2 - 4x^2

and its tangent line

y = -8x + 6

meeting exactly at the point of tangency

(1, -2)

Exercises

Find the equation of the tangent line to the curve $y = x^2 - 6\frac{1}{5}x + 14\frac{1}{2}$ which is parallel to the line $x + 2y + 3 = 0$ .
Determine the equation of the tangent line to the parabola $y = 2x^2 - 3x + 5$ at the point with an ordinate of $4$ .
A curve $y = 3x - \frac{3}{x^2}$ intersects the $x$ -axis at P. Find the equation of the tangent line to the curve at point $P$ !
The curve $y = (x^2 + 2)^2$ intersects the $y$ -axis at point $A$ . Show that the tangent line to the curve at point $A$ is parallel to the $x$ -axis and is $4 \text{ units}$ away from the origin!
Determine the coordinates of the point on the curve $y = 2x^2 - 7x + 1$ , if the tangent line to the curve at that point forms an angle of $45^\circ$ with the positive $x$ -axis. Also, determine the equation of the tangent line to the curve that passes through that point!

Answer Key

Solution:

Step $1$ : Determine the gradient.

The tangent line must be parallel to the line $x + 2y + 3 = 0$ . To find its gradient, let's first convert this line's equation into the slope-intercept form $y = mx + c$ .
$2y = -x - 3$
From this, we know the gradient of the line is $m = -\frac{1}{2}$ . Since the tangent line is parallel, its gradient is the same.

Step $2$ : Find the point of tangency.

The gradient of the curve at a point is equal to the value of the first derivative at that point. First, let's find the derivative function of $f(x) = x^2 - \frac{31}{5}x + \frac{29}{2}$ .

$f'(x) = 2x - \frac{31}{5}$

Next, we set this derivative equal to the gradient we know ( $m = -1/2$ ) to find the $x$ -coordinate of the point of tangency.
$2x - \frac{31}{5} = -\frac{1}{2}$
After getting the $x$ -coordinate $x=\frac{57}{20}$ , we substitute this value back into the original curve equation to find its $y$ -coordinate.

$y = (\frac{57}{20})^2 - \frac{31}{5}(\frac{57}{20}) + \frac{29}{2} = \frac{3249}{400} - \frac{1767}{100} + \frac{29}{2} = \frac{3249 - 7068 + 5800}{400} = \frac{1981}{400}$

So, the point of tangency is $\left(\frac{57}{20}, \frac{1981}{400}\right)$ .

Step $3$ : Construct the line equation.

With the point $\left(\frac{57}{20}, \frac{1981}{400}\right)$ and gradient $m = -\frac{1}{2}$ , the equation is:
$y - \frac{1981}{400} = -\frac{1}{2}\left(x - \frac{57}{20}\right)$
Solution:

Step $1$ : Find the point(s) of tangency.

Since the ordinate is $4$ , we set $y=4$ in the parabola's equation.
$4 = 2x^2 - 3x + 5$
From factorization, we get two x-values: $x=1$ and $x=\frac{1}{2}$ . This means there are two points of tangency: $(1, 4)$ and $\left(\frac{1}{2}, 4\right)$ . Therefore, there will be two tangent line equations.

Step $2$ : Calculate the gradient and create the equation for each point.

We will process each point of tangency separately. The derivative of the function is $f'(x) = 4x - 3$ .

Case One: Point $(1, 4)$

The gradient at this point is $m = f'(1) = 4(1) - 3 = 1$ .

Thus, the equation is:

$y - 4 = 1(x - 1) \implies y = x + 3$

Case Two: Point $\left(\frac{1}{2}, 4\right)$

The gradient at this point is $m = f'(\frac{1}{2}) = 4(\frac{1}{2}) - 3 = -1$ .

Thus, the equation is:

$y - 4 = -1(x - \frac{1}{2}) \implies y = -x + \frac{9}{2}$
Solution:

Step $1$ : Find point $P$ .

The curve intersects the $x$ -axis when $y=0$ .
$0 = 3x - \frac{3}{x^2}$
So, the intersection point $P$ is $(1, 0)$ .

Step $2$ : Find the gradient at P.

Let's first rewrite the function as $f(x) = 3x - 3x^{-2}$ to make it easier to differentiate.

$f'(x) = 3 - (-2)(3)x^{-3} = 3 + \frac{6}{x^3}$

The gradient at $x=1$ is $m = f'(1) = 3 + \frac{6}{1^3} = 9$ .

Step $3$ : Construct the equation.

With point $(1, 0)$ and gradient $9$ , the equation is:

$y - 0 = 9(x - 1) \implies y = 9x - 9$
Solution:

Step $1$ : Find point $A$ .

The curve intersects the $y$ -axis when $x=0$ .

$y = (0^2 + 2)^2 = 4$

So, the intersection point $A$ is $(0, 4)$ .

Step $2$ : Prove the tangent line is parallel to the $x$ -axis.

A line parallel to the $x$ -axis must have a gradient of $0$ . Let's prove that the derivative of the function at point $A$ ( $x=0$ ) is zero.

The derivative of $f(x)=(x^2+2)^2$ using the chain rule is $f'(x) = 2(x^2+2)(2x) = 4x(x^2+2)$ .

The gradient at $x=0$ is $m = f'(0) = 4(0)(0^2+2) = 0$ .

Since the gradient is zero, it is proven that the tangent line is parallel to the $x$ -axis.

Step $3$ : Prove its distance is $4 \text{ units}$ from the origin.

The equation of the tangent line at point $(0,4)$ with gradient $m=0$ is:

$y - 4 = 0(x - 0) \implies y = 4$

The line $y=4$ is a horizontal line. The distance from any point on this line to the $x$ -axis (the line $y=0$ ) is $4 \text{ units}$ . Since the origin $(0,0)$ lies on the $x$ -axis, the distance from this tangent line to the origin is also $4 \text{ units}$ . Proven.
Solution:

Step $1$ : Determine the gradient from the angle.

The relationship between the gradient ( $m$ ) and the angle ( $\theta$ ) a line makes with the positive $x$ -axis is given by $m = \tan(\theta)$ .

$m = \tan(45^\circ) = 1$

So, the gradient of the tangent line we are looking for is $1$ .

Step $2$ : Find the coordinates of the point of tangency.

The gradient is also the first derivative of the curve $f(x) = 2x^2 - 7x + 1$ .

$f'(x) = 4x - 7$

We set it equal to the gradient we found:
$4x - 7 = 1$
Now, find the $y$ -value by plugging $x=2$ into the curve's equation:

$y = 2(2)^2 - 7(2) + 1 = 8 - 14 + 1 = -5$

So, the coordinates of the point of tangency are $(2, -5)$ .

Step $3$ : Determine the equation of the tangent line.

Using the point $(2, -5)$ and gradient $m=1$ :
$y - (-5) = 1(x - 2)$